1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Harm osc, anharmonic perturbation

  1. Jan 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Particle bound by
    [tex] V = \frac{1}{2} m \omega^2 x^2 - a x^3 [/tex]
    for small x. Show that the mean position of the particle changes with the energy of the eigenstates when [tex]a[/tex] is small, so first order perturbation theory works.

    2. Relevant equations
    For the harmonic oscillator
    [tex] x = \sqrt{\frac{\hbar}{2m\omega}}(a^{\dagger}+a) [/tex]

    3. The attempt at a solution
    That x^3 perturbation will give an odd number of creation/destruction operators, so there's no shift in energy eigenvalues to first order in perturbation theory. But how does that help answering the question?
  2. jcsd
  3. Jan 2, 2008 #2

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    There is a first order transition, but it is between different energy states.
  4. Jan 4, 2008 #3
    But how does that help answering the question?
  5. Jan 12, 2008 #4
    Noone have any ideas on this?
  6. Jan 22, 2008 #5
    I'm not going to give up on this! The question asks to calculate the mean position of the particle, so if I do that:
    [tex]<x> = \sqrt{\frac{\hbar}{2m\omega}}<n|a^{\dagger}+a|n> = 0 [/tex]
    That's just zero, haveing nothing to do with energy eigernstates or size of a?

    Sure I could calculate the perturbated energy to second order, getting terms like
    [tex] a^2 \big( \frac{\hbar}{2m\omega} \big)^3 <n|a^{\dagger} aa \frac{1}{E_n-H_0} a^{\dagger} a a^{\dagger} |n> = -n(n+1)^2 \frac{\hbar^2 a^2}{8m^3 \omega^4}[/tex]
    and so on, but I don't see how that has anyhting to do with the question asked.
  7. Jan 22, 2008 #6
    Shouldn't you consider the expanded eigenkets?

    [tex] | n > = |n^{(0)}>+\lambda |n^{(1)}> + \lambda^2 |n^{(2)}> + \dots[/tex]

    Then you get to first order something like:

    [tex] <x> = < n^{(0)} | x | n^{(0)} > + \lambda \left( < n^{(0)} | x | n^{(1)} > + < n^{(1)} | x | n^{(0)} > \right)+ \mathcal{O} (\lambda^2) = \lambda \left( < n^{(0)} | x | n^{(1)} > + < n^{(1)} | x | n^{(0)} > \right)+ \mathcal{O} (\lambda^2)[/tex]

    Could that be the correct approach?
    Last edited: Jan 22, 2008
  8. Jan 22, 2008 #7
    I think that you dont really have to solve for the corrected energy's. You need the new eigenkets, and then just solve for the expectation value, <x>, by taking the inner product of x and the eigenkets.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Harm osc, anharmonic perturbation
  1. Anharmonic Oscillation (Replies: 4)